\section{The Keplerian two-body problem}

\begin{frame}[t]{\thesection.\ \insertsection}
    We are going to look at the Keplerian two-body problem.
    \newline
    \begin{itemize}
        \item The orbital problem for \textcolor{blue}{two point masses}
        \begin{itemize}
            \item[\mysquare]\textcolor{blue}{Treating the planet and spacecraft as point masses} provides a great
            deal of insight into the behavior of a spacecraft in orbit about a planet.
        \end{itemize}
        \item It turns out that this is a reasonable model for spacecraft orbiting planets, planets orbiting the sun, and others.
    \end{itemize}
    \vspace{12pt}
    \centering\includegraphics[scale=1.2]{fig_2_1.pdf} \\
    \textcolor{blue}{Figure \arabic{section}.1:} Two-body problem
\end{frame}

\begin{frame}[t]{\thesection.\ \insertsection}
    The Newton laws only hold in an inertial frame of reference.
    We can only find approximate inertial frames, for example,
    \begin{block}{Heliocentric Coordinate System}
        Origin: at the center of the Sun \\
        Orientation: fixed with respect to the stars
    \end{block}
    \begin{center}\includegraphics[scale=0.8]{fig_2_2.pdf}\end{center}
    \begin{center}\textcolor{blue}{Figure \arabic{section}.2:} Heliocentric coordinate system\end{center}
\end{frame}

\begin{frame}[t]{\thesection.\ \insertsection}
\begin{block}{Property of Inertial frames}
    Any reference frame that satisfies
    \begin{enumerate}
        \item Not rotating
        \item The origin being either not translating, or translating with constant velocity
    \end{enumerate}
    With respect to an original inertial reference frame is also inertial.
\end{block}
By simple calculation, the angular velocity of the Earth with respect to
the Sun: $360^\circ/(365\times 24) = 0.0417^\circ/\text{hour}$
\begin{columns}
\column{0.5\textwidth}
\begin{block}{Earth-Centered-Inertial (ECI) frame}
    Origin: at the center of the Earth \\
    Orientation: fixed with respect to the stars
\end{block}
\column{0.4\textwidth}
\vspace{-12pt}
\begin{adjustwidth}{-3pt}{-20pt}
    \begin{center}\includegraphics{fig_2_3.pdf}\end{center}
    \hspace{-8pt}\textcolor{blue}{Figure \arabic{section}.3:} Earth-Centered-Inertial (ECI) frame
\end{adjustwidth}
\end{columns}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Equations of motion}
\begin{frame}[t]{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
    Consider two point masses of mass $m_1$ and $m_2$. \\
    \centering\includegraphics{fig_2_1.pdf} \\
    \textcolor{blue}{Figure \arabic{section}.4:} Two-body problem
\vspace{10pt}

    \begin{itemize}
        \item In order to apply Newton’s laws of motion,
            we require an inertial reference frame $\mathcal F_1$.
        \item The equation of motion for the position of $m_2$ with respect to $m_1$ is
        \[\ddot{\vec{_{\,}r}}_{21} = -\frac{G(m_1+m_2)}{|\vec{_{}r}_{21}|^3}\vec{_{}r}_{21}\]
        where $\vec{_{}r}_{21}=\vec{_{}r}_{2}-\vec{_{}r}_{1}$ is the position of mass $m_2$ with respect to mass $m_1$,
        and $G$ is Newton’s universal gravitational constant.
    \end{itemize}
\end{frame}

\begin{frame}[t]{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
    Now, we will assume that $m_1 >> m_2$.
    \begin{itemize}
        \item As is the case for example when $m_1$ is a planet and $m_2$ is a spacecraft,
        or when $m_1$ is the sun and $m_2$ is a planet.
    \end{itemize}
    Under this assumption, $m_1 + m_2 \approx m_1$, and
    \[\ddot{\vec{_{\,}r}}_{21} = -\frac{G m_1}{|\vec{_{}r}_{21}|^3}\vec{_{}r}_{21}\]
    \begin{block}{The Keplerian two-body orbital equation of motion}
        \[\ddot{\vec{_{\,}r}}_{21} = -\frac{\mu}{|\vec{_{}r}_{21}|^3}\vec{_{}r}_{21}\]
        where the gravitational constant $\mu = Gm_1$.
    \end{block}
    \textcolor{blue}{The Earth’s gravitational constant} $3.986 \times 10^{5}\text{km}^3/\text{s}^2$ \\
    \textcolor{blue}{The Sun’s gravitational constant} $1.3271244 \times 10^{11}\text{km}^3/\text{s}^2$
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Constants of the motion}
\begin{frame}[t]{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
1. Orbital augular momentum
\begin{columns}
\column{0.6\textwidth}
    \begin{block}{Definition \arabic{section}.1}
        The orbital augular momentum is defined as
        \[\vec h \triangleq \vec{_{}r} \times \vec v\]
    \end{block}
\column{0.3\textwidth}
\centering\includegraphics{fig_2_p6.pdf} \\
\end{columns}
\vspace{10pt}
We can prove that
\[ \dot{\vec h}=\vec 0 \]
\begin{itemize}
    \item $\vec h$ is constant with respect to $\mathcal F_I$.
    \item By $\vec{_{}r}\perp\vec h$, $\vec{_{}r}$
        \textcolor{blue}{evolves on a plane in inertial space} that is perpendicular to $\vec h$.
\end{itemize}
\end{frame}

\begin{frame}[t]{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
2. Orbital energy
\begin{block}{Definition \arabic{section}.2}
    The orbital energy is defined as
    \vspace{-12pt}
    \[\mathcal E\triangleq\frac{v^2}{2}-\frac{\mu}{r} \vspace{-6pt} \]
    where $v = |\vec{_{}v}|$ is the orbital speed, and $r = |\vec{_{}r}|$ is the distance of $m_2$ from $m_1$.
\end{block}
\vspace{-21pt}
\begin{align*}
    \textcolor{blue}{\frac{v^2}{2}}  & \text{: the kinetic energy per unit mass} \\
    \textcolor{blue}{-\frac{\mu}{r}} & \text{: the two-body gravitational potential energy}
\end{align*}
\vspace{-6pt}
We can prove that
\[\dot{\mathcal E}=0\]
which means that the two-body orbital energy is constant.
\end{frame}

\begin{frame}[t]{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
3. The eccentricity
\begin{block}{Definition \arabic{section}.3}
    The eccentricity is defined as
    \[\vec e\triangleq\frac{\vec v\times\vec h}{\mu}-\frac{\vec{_{}r}}{r}\]
\end{block}
\begin{center}\includegraphics{fig_2_p6.pdf}\end{center}
We can prove that
\[\dot{\vec e}=\vec 0\]
which means that the eccentricity vector $\vec e$ lies in the orbital plane, and is constant.
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Shape of a Keplerian orbit}

\begin{frame}{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
    \begin{itemize}\setlength{\itemsep}{20pt}
        \item We have established that the Keplerian orbital motion is planar, and we can specify the orbital plane by the angular momentum vector $\vec h$.
        \item We have also found a vector that is fixed in the orbital plane, that is the eccentricity vector $\vec e$.
    \end{itemize}
\end{frame}

\begin{frame}[t]{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
    We can obtain
    \[r=\frac{p}{1+e\cos\theta}\]
    where $p=h^2/\mu$.
    \vspace{20pt}
    \begin{center}\includegraphics{fig_2_5.pdf}\end{center}
    \begin{center}\textcolor{blue}{Figure \arabic{section}.5:} conic curve\end{center}
    
\end{frame}

\begin{frame}[t]{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
    We have the relationship among the orbital energy $\mathcal E$, the eccentricity $e$, and the angular momentum $h$:
    \[\mathcal E=\frac{v^2}{2}-\frac{\mu}{r}=\frac{\mu^2(e^2-1)}{2h^2},
    v=\sqrt{\frac{2\mu}{r}+2\mathcal E}\]
    From this, we see that
    \color{blue}\begin{alignat*}{2}
        &\text{Hyperbola} &\quad& e > 1\Rightarrow\mathcal E > 0 \\
        &\text{Parabola}  &\quad& e = 1\Rightarrow\mathcal E = 0 \\
        &\text{Ellipse}   &\quad& e < 1\Rightarrow\mathcal E < 0 \\
    \end{alignat*}\color{black}
    For example, let us consider the height $h = 3.5786\times 10^7\text{m}$.
    For a circular orbit, the velocity should satisfy:
    \[v<\sqrt{\frac{2\mu}{r}}=4.3482\times10^3\text{m/s}\]
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Kepler's Laws}

\begin{frame}[t]{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
    Johannes Kepler postulated the following laws based on empirical data obtained by Tycho Brahe. \\
    \begin{center}\includegraphics[scale=0.2]{fig_2_6.jpg}\end{center}
    \begin{center}\textcolor{blue}{Figure \arabic{section}.6:} Johannes Kepler\end{center}
\end{frame}

\begin{frame}[t]{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
\begin{block}{Kepler's Laws}
\begin{enumerate}
    \item The orbit of each planet is an ellipse with the sun at one focus.
    \begin{center}\includegraphics{fig_2_7_1.pdf} \quad
    \includegraphics{fig_2_7_2.pdf}\end{center}
    \begin{center}\textcolor{blue}{Figure \arabic{section}.7:} Keplerian laws\end{center}
    \item The radius vector drawn from the sun to the planet sweeps out equal areas in equal times.
    \item The squares of the periods of the planetary orbits are proportional to the cubes of the semi-major axes of the orbits:
    \[ T^2 = \frac{4\pi^2}{\mu} a^3 \]
\end{enumerate}
\end{block}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Time of flight}

\begin{frame}[t]{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
We need to determine how \( m_2 \) traverses the orbit with respect to time, that is, \(\theta(t)\). \\
1. Circular Orbit
\begin{center}\includegraphics{fig_2_p15.pdf}\end{center}
\[ \theta(t) = \theta(t_0) + \sqrt{\frac{\mu}{r^3}} (t - t_0) \]
where
\begin{itemize}
    \item Earth's gravitational constant \(\mu_E = 3.986 \times 10^5 \, \text{km}^3/\text{s}^2\)
    \item Sun's gravitational constant \(\mu_S = 1.3271244 \times 10^{11} \, \text{km}^3/\text{s}^2\)
\end{itemize}
\end{frame}

\begin{frame}[t]{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
2. Elliptical orbits
\begin{center}\includegraphics[scale=0.9]{fig_2_p16.pdf}\end{center}
Let \( t_0 \) be the time of periapsis passage, that is \(\theta(t_0) = 0\). \\
$t \Rightarrow \theta(t):$
\begin{enumerate}
    \item \( M = \sqrt{\frac{\mu}{a^3}}(t - t_0) \)
    \item Solve Kepler's equation \( E - e \sin E = M \) for \( E \)
    \item \( \tan \frac{\theta}{2} = \sqrt{\frac{1+e}{1-e}} \tan \frac{E}{2} \)
\end{enumerate}
\end{frame}

\begin{frame}[t]{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
\begin{columns}
\column{0.45\textwidth}
For the Kepler's equation
\[ E - e \sin E = M \]
we can use an iterative scheme to solve \( E \) given \( M \):
\column{0.45\textwidth}
\begin{center}\includegraphics{fig_2_p17.pdf}\end{center}
\end{columns}
\end{frame}

\begin{frame}[t]{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
3. Parabolic orbits
\begin{center}\includegraphics{fig_2_p18.pdf}\end{center}
\[ \tan^3 \frac{\theta}{2} + 3 \tan \frac{\theta}{2} = 6 \sqrt{\frac{\mu}{p^3}} (t - t_0) \]
\end{frame}

\begin{frame}[t]{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
4. Hyperbolic orbits
\begin{enumerate}
    \item \( N = \sqrt{\frac{\mu}{-a^3}} (t - t_0) \)
    \item Solve Kepler's equation \[e\sinh H - H = N\] for \( H \)
        using the similar iterative method as for $E$.
    \item \( \tan \frac{\theta}{2} = \sqrt{\frac{e+1}{e-1}} \tanh \frac{H}{2} \)
\end{enumerate}
where
\begin{align*}
    &\sinh x = \frac{e^x - e^{-x}}{2} \\
    &\cosh x = \frac{e^x + e^{-x}}{2} \\
    &\tanh x = \frac{\sinh x}{\cosh x}
\end{align*}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Orbital elements}

\begin{frame}[t]{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
\vspace{-15pt}
\begin{center}\includegraphics[scale=0.8]{fig_2_8.pdf}\end{center}
\vspace{-18pt}
\begin{center}\textcolor{blue}{Figure \arabic{section}.8:} Orbital elements\end{center}
\vspace{-15pt}
\begin{block}{Geocentric-Equatorial Coordinate System $\mathcal F_G$ (\( \vec{X}_G \vec{Y}_G \vec{Z}_G \)) (Inertial)}
\vspace{-15pt}
\begin{align*}
    \textcolor{blue}{\text{Origin}}\quad & \text{at the center of the Earth} \\
    \textcolor{blue}{\text{Orientation}}\quad & \vec{X}_G \text{: points towards the vernal equinox} \\
    & \vec{Z}_G \text{: points towards the north pole} \\
    & \vec{Y}_G \text{: completes the right-handed coordinate system}
\end{align*}\end{block}
\end{frame}

\begin{frame}[t]{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
\vspace{-4pt}
\begin{center}\includegraphics[scale=0.6]{fig_2_8.pdf}\end{center}
\vspace{-10pt}
\begin{enumerate}
    \item \( i \) - inclination: the angle between \( \vec{h} \) and \( \vec{Z}_G \)
    \begin{itemize}
        \item[\mysquare] to specify the orbital plane
    \end{itemize}
    \item \( \Omega \) - right ascension of the ascending node
    \begin{itemize}
        \item[\mysquare] to specify the orientation of the orbit
        \item[\mysquare] The line of nodes \( \vec{n} \): the line of intersection between the equatorial and orbital planes
    \end{itemize}
    \item \( \omega \) - argument of perigee: the angle between \( \vec{n} \) and \( \vec{e} \)
    \begin{itemize}
        \item[\mysquare] to specify the orientation of the eccentricity $\vec e$
    \end{itemize}
\end{enumerate}
\vspace{-4pt}
\begin{columns}
\column{0.3\textwidth}
\begin{adjustwidth}{-18pt}{0pt}\begin{enumerate}
    \setcounter{enumi}{3}
    \item \( a \) - semi-major axis
\end{enumerate}\end{adjustwidth}
\column{0.3\textwidth}
\begin{adjustwidth}{-27pt}{0pt}\begin{enumerate}
    \setcounter{enumi}{4}
    \item \( e \) - eccentricity
\end{enumerate}\end{adjustwidth}
\column{0.3\textwidth}
\begin{adjustwidth}{-50pt}{-10pt}\begin{enumerate}
    \setcounter{enumi}{5}
    \item \( t_0 \) - time of perigee passage
\end{enumerate}\end{adjustwidth}
\end{columns}
\end{frame}

\begin{frame}{\thesection.\ \insertsection \\ \small\thesection.\thesubsection\ \insertsubsection}
The inertial position \( \vec{_{}r} \) and velocity \( \vec{v} \) at time \( t \) \(\Leftrightarrow\) Orbital elements \\
\vspace{16pt}
Reference: \\
A. H. J. de Ruiter, C. J. Damaren, J. R. Forbes, \textit{Spacecraft Dynamics and Control, an Introduction}, John Wiley and Sons Ltd, 2013.
\end{frame}
